Using Newtons Method to Solve Equations | Homework Help

[Asphyxiated]

Homework Statement

Solve each equation using Newtons method. The problem I am working on right now is:

x^{5}+x^{3}+x=1

which is the same as:

x^{5}+x^{3}+x-1=0

Homework Equations

x_{2} = x_{1}-\frac {f(x)}{f'(x)}

with x₁ being the first guess to what an f(x)=0 would be.

The Attempt at a Solution

So I put the equation into my calculator to graph it and It is very clear from the graph that f(x)=0 on the x>0 side so I choose x₁=1 and continue like so:

x_{2} = 1 - \frac {f(x)}{f'(x)} = 1- \frac {2}{9} = \frac {7}{9}

and after 5 iterations I get:

x_{5} = .6368843716

which looks right to me but when I check the answer in the back of the book it says that it should be -.63688, so just the opposite of what I got. The graph of the equation clearly shows that f(x)=0 on x>0 so did I do something wrong or does the book have another typo?

thanks

 

[Char. Limit]

Another typo.

 

[hunt_mat]

I've just spent the last 5 months play with Newton's method... The general method is
<br /> x_{n+1}=x_{n}-\frac{f(x_{n})}{f&#039;(x_{n})}<br />
Where in your case:
<br /> f(x)=x^{5}+x^{3}+x-1<br />
The derivative of this is:
<br /> f&#039;(x)=5x^{4}+3x^{2}+1<br />
And in your case Newton's method is
<br /> x_{n+1}=x_{n}-\frac{x_{n}^{5}+x_{n}^{3}+x_{n}-1}{5x_{n}^{4}+3x_{n}^{2}+1}<br />
So all you have to do is plug numbers in. Newtons method converges very quickly in my experience, 2 or 3 iterations will usually do the trick.

 

[hunt_mat]

As a check, plug your solution back into the equation and check.